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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    A 120 V AC source is connected across a pure inductor of inductance 0.70 H. If the frequency of the source is 60 Hz, the current passing through the inductor is

    A)  4.55 A                                  

    B)  0.355 A

    C)  0.455 A                               

    D)  3.55 A

    Correct Answer: C

    Solution :

    Given,   \[{{V}_{L}}=120volt,\] \[L=0.70H,\] \[f=60Hz\] \[\therefore \]current passing through inductor                 \[{{I}_{L}}=\frac{{{V}_{L}}}{{{X}_{L}}}=\frac{{{V}_{L}}}{\omega L}=\frac{{{V}_{L}}}{2\pi fL}\]                 \[=\frac{120}{2\pi \times 60\times 0.70}=0.455A\]


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